Question

The temperature of a liquid falls from 365 K to 359 K in 3 minutes. The time during which temperature of this liquid falls from 342 K to 338 K is [Let the room temperature be 296 K]

• A. 6 min
• B. 4 min
• C. 3 min
• D. 2 min

Answer 1

Answer: (C)

3 min

Answer 2

Using Newton’s law of cooling, the rate of temperature change is given by

$$\frac{dT}{dt} = -K\,(T-T_{\text{room}})$$

For a small temperature drop ΔT over a short interval, we approximate

$$\Delta T \approx K\,(T_{\text{avg}}-T_{\text{room}})\,\Delta t.$$

Interval 1:

• Temperature drops from 365 K to 359 K: $\Delta T_1 = 365 - 359 = 6\,K.$
• Time: $\Delta t_1 = 3\,\text{min}.$
• Average temperature: $T_{\text{avg1}} = \frac{365+359}{2} = 362\,K.$
• Thus, $$6 = K\,(362-296)\,(3) \quad \Rightarrow \quad 6 = 3K\,(66).$$ Solving, $$K = \frac{6}{198} = \frac{1}{33}\, \text{min}^{-1}.$$

Interval 2:

• Temperature drops from 342 K to 338 K: $\Delta T_2 = 342 - 338 = 4\,K.$
• Average temperature: $T_{\text{avg2}} = \frac{342+338}{2} = 340\,K.$
• Let the time needed be $t$. Then, $$4 = K\,(340-296)\,t = \frac{1}{33}\times 44 \, t.$$ Simplify: $$\frac{44}{33}\,t = \frac{4}{1} \quad \Rightarrow \quad \frac{4}{3}\,t = 4.$$ Therefore, $$t = 4 \times \frac{3}{4} = 3\, \text{min}.$$